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Dmitry [639]
3 years ago
14

What direction would the north pole of a bar magnet point if you were to hang the bar magnet from a thin string?.

Physics
1 answer:
bearhunter [10]3 years ago
3 0

Answer:

TOWARD the North pole.

Explanation:

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A helium tank holds a volume of .02 m^3 at a pressure of 15.5*10^6 Pa and a temperature of 293 K. How many spherical balloons wi
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Answer:

Explanation:

We shall find the volume of stored gas at atmospheric pressure .

P₁ V₁ = P₂ V₂

15.5 x 10⁶ x .02 = 10⁵ x V₂  ( atmospheric pressure is 10⁵ Pa )

V₂ = 3.1 m³

volume of one balloon = 4/3 x π r³ , r is radius of balloon

= 4/3  π x .11³

= .050658 m³

no of balloon = total volume to be filled / volume of one balloon

= 3.1 / .050658

= 61 .2

= 61

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3 years ago
How does Newton's second law of motion can be used to calculate the acceleration of an object?
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Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

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4 years ago
A student performs an activity using a sheet of paper, a magnet, and a steel ball. The image shows the setup. The student observ
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Answer:b

Explanation:

6 0
3 years ago
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In order for work to happen you MUST have?
Nataly_w [17]

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7 0
3 years ago
What is the maximum value of the magnetic field at a<br> distance2.5m from a 100-W light bulb?
MA_775_DIABLO [31]

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

I = \frac{P}{A}

I = \frac{P}{4\pi r^2}

Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that

I = \frac{100}{4\pi (2.5)^2}

I = 1.2738W/m^2

The relation between intensity I and E_{max}

I = \frac{E_max^2}{2\mu_0 c}

Here,

\mu_0 = Permeability constant

c = Speed of light

Rearranging for the Maximum Energy and substituting we have then,

E_{max}^2 = 2I\mu_0 c

E_{max}=\sqrt{2I\mu_0 c }

E_{max} = 2(1.2738)(4\pi*10^{-7})(3*10^8)

E_{max} = 30.982 V/m

Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,

B_{max} = \frac{E_{max}}{c}

B_{max} = \frac{30.982 V /m}{3*10^8}

B_{max} = 1.03275 *10{-7} T

Therefore the maximum value of the magnetic field is B_{max} = 1.03275 *10{-7} T

3 0
3 years ago
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